Using integration by parts to solve

Question

This question has been asked in a similar format, but not that I've found exactly the same as mine. It may be that they are the same, but I don't understand a fundamental part so would appreciate some guidance.

I am trying to solve this:

$\int_{0}^{\infty}{ \lambda x e^{- \lambda x}} dx$

I have done integration by parts once using $u=x$ and $dv = e^{-\lambda x}$ after taking the $\lambda$ outside the integral, thus getting:

$\lambda$ $( - \dfrac{x e^{- \lambda x}}{ \lambda }$ $\Biggr|_{0}^{\infty}$ $-$ $\int_{0}^{\infty}{\dfrac{e^{-\lambda x}}{\lambda}}dx)$

and simplifying:

$( - x e^{- \lambda x}$ $\Biggr|_{0}^{\infty}$ $-$ $\int_{0}^{\infty}{e^{-\lambda x}}dx)$

Then taking the integral of the second part substituting $u=-\lambda x$ and $du=-\lambda dx$:

$- \dfrac{1}{\lambda e^{\lambda x}}$

From this point I now have:

$- x e^{- \lambda x}$ $\Biggr|_{0}^{\infty}$ $- \dfrac{1}{\lambda e^{\lambda x}}$

What I am confused about from here is that when I take this limits for the first part I get infinity, and then I have no limits to take for the second part any more. I know the answer is $\dfrac{1}{\lambda}$ but I am unsure how to get here.

I have found this question which appears to be asking the same thing: Finding $\int_0^{\infty}xe^{-\lambda x} \, dx$

However I still don't understand how I get the $\dfrac{1}{\lambda}$.

Thank you in advance.

Answer 1

For this case after working out the integral should be:

$-xe^{-\lambda x}-\dfrac{e^{-\lambda x}}{\lambda}$

$=-\dfrac{1}{e^{\lambda x}} \left(x+\dfrac{1}{\lambda} \right) \Biggr|_{0}^{\infty}$

The upper bound gives a value tending to $0$ because the denominator of $\dfrac{1}{e^{\lambda x}}$ increases without bound. At $0$ we have:

$=-\dfrac{1}{e^{\lambda \cdot 0}} \left(0+\dfrac{1}{\lambda} \right)=-\dfrac{1}{\lambda}$

So the result is $0-\left(-\dfrac{1}{\lambda} \right)=\dfrac{1}{\lambda}$

Maybe it's the way you did the integral. Doing the integration by parts should have given:

$I(x)=\lambda x \cdot -\dfrac{e^{-\lambda x}}{\lambda} - \displaystyle \int \lambda \cdot -\dfrac{e^{-\lambda x}}{\lambda}$

$=-xe^{-\lambda x}+\displaystyle \int e^{-\lambda x}$

$=-xe^{-\lambda x}-\dfrac{e^{-\lambda x}}{\lambda}$

$=\color{blue}{\boxed{-e^{-\lambda x}\left(x+ \dfrac{1}{\lambda} \right)}}$

Edit: The integration by parts section was a little wrong. All good now.

Answer 2

With $z:=\lambda x$,

$$\int_0^\infty \lambda xe^{-\lambda x}dx=\frac1\lambda\int_0^\infty ze^{-z}.$$

At this stage, you already know that if the integral converges, the answer will be of the form $\dfrac c\lambda$ !

By parts,

$$\int ze^{-z}dz=-ze^{-z}+\int e^{-z}dz=-ze^{-z}-e^{-z}.$$

Hence with the given integration limits, $c=1$.

Answer 3

The left hand part of your integral evaluates as zero because $\frac{x}{e^x} \rightarrow 0$ for $x \rightarrow \infty$. Continuing with the substitution that you are carrying out: $$u = -\lambda x$$ $$du = -\lambda dx$$ so $dx = -1/\lambda du$. Finally \begin{align} -\int_{0}^{\infty}{\dfrac{e^{-\lambda x}}{\lambda}}dx&=-\int_{0}^{\infty}{\dfrac{e^{u}}{\lambda}}du\\ &= -\dfrac{e^{u}}{\lambda} \Biggr|_{-\infty}^{0}\\ &= -(0-\dfrac{1}{\lambda}) = \dfrac{1}{\lambda} \end{align}

Answer 4

You don't actually bound $\lambda$, but assuming $\lambda > 0$, $$ \lim_{x \rightarrow \infty} -x \mathrm{e}^{\lambda x} = 0 \text{,} $$ by many methods, including l'Hopital's rule.

Then, it is baffling that you find $\frac{-1}{\lambda \mathrm{e}^{-\lambda x}}$ as an antiderivative for evaluating $\int_0^\infty \mathrm{e}^{-\lambda x}\,\mathrm{d}x$, because both this is not an antiderivative of $\mathrm{e}^{-\lambda x}$ and you do not evaluate the antiderivative at the limits of the interval of integration (by a limit at the improper end). A correct antiderivative is $\frac{-\mathrm{e}^{-\lambda x}}{\lambda}$.